On Nonuniformly Subelliptic Equations of Q sub-laplacian Type with Critical Growth in the Heisenberg Group

Transcription

1 Advanced Nonlinear Studies ), On Nonuniformly Subelliptic Equations of sub-laplacian Type with Critical Growth in the Heisenberg Group Nguyen Lam, Guozhen Lu Department of Mathematics Wayne State University, Detroit, Michigan Hanli Tang School of Mathematical Sciences Beijing Normal University, Beijing, China Received in revised form 5 April 212 Communicated by Susanna Terracini Abstract Let H n = R 2n R be the n dimensional Heisenberg group, H n be its subelliptic gradient operator, and ρ ξ) = z 4 + t 2) 1/4 for ξ = z, t) H n be the distance function in H n Denote H = H n, = 2n + 2 and = / 1) Let be a bounded domain with smooth boundary in H Motivated by the Moser-Trudinger inequalities on the Heisenberg group, we study the existence of solution to a nonuniformly subelliptic equation of the form div H a ξ, H u ξ))) = f ξ,uξ)) ρξ) u W 1, )\ {} u in + εhξ) in where f : R R behaves like exp α u ) when u In the case of sub- Laplacian div H H u 2 H u ) = f ξ,u) + εhξ) in ρξ) u W 1, )\ {}, u in we will apply minimax methods to obtain multiplicity of weak solutions 21 Mathematics Subject Classification 42B37, 35J92, 35J62 Key words Moser-Trudinger inequality, Heisenberg group, subelliptic equations, -sublaplacian, Mountain-Pass theorem, Palais-Smale sequences, existence and multiplicity of solutions Research is partly supported by a US NSF grant DMS Corresponding Author: Guozhen Lu at gzlu@mathwayneedu, 659

2 66 N Lam, G Lu, H Tang 1 Introduction Let R n be an open and bounded set and W 1,p ) n 2) be the completion of C ) under the norm 1/p u 1,p W ) := u p dx + u p dx Then the Sobolev embeddings say that W 1,p ) L np n p ) if 1 p < n W 1,p ) C 1 n p ) if n < p The case p = n can be seen as the limit case of these embeddings and it is known that W 1,n ) L q ) for 1 q < However, by some easy examples, we can conclude that W 1,n ) L ) It is showed by Judovich [18], Pohozaev [38] and Trudinger [43] independently that W 1,n ) L ϕn ) where L ϕn ) is the Orlicz space associated with the Young function ϕ n t) = exp t n/n 1)) 1 Extending this result, J Moser [37] finds the largest positive real number n = nω 1 n 1 n 1, where ω n 1 is the area of the surface of the unit n ball, such that if is a domain with finite n measure in Euclidean n space R n, n 2, then there is a constant c depending only on n such that 1 n exp ) u n 1 dx c for any n, any u W 1,n ) with u n dx 1 Moreover, this constant n is sharp in the meaning that if > n, then the above inequality can no longer hold with some c independent of u Such an inequality is nowadays known as Moser-Trudinger type inequality However, when has infinite volume, the result of J Moser is meaningless In this case, the sharp Moser-Trudinger type inequality was obtained by B Ruf [39] in dimension two and YX Li-Ruf [29] in general dimension The Moser-Trudinger type inequalities has been extended to many different settings: high order derivatives by D Adams which is now called Adams type inequalities [1, 2, 23, 35, 4, 42]; compact Riemannian manifolds without boundary by Fontana [17] see [28]); singular Moser-Trudinger inequalities which are the combinations of the Hardy inequalities and Moser-Trudinger inequalities are established in [3, 5, 23] It is also worthy to note that the Moser-Trudinger type inequalities play an essential role in geometric analysis and in the study of the exponential growth partial differential equations where, roughly speaking, the nonlinearity behaves like e α u n n 1 as u Here we mention Atkinson-Peletier [6], Carleson-Chang [8], Flucher [16], Lin [3], Adimurthi et al [2, 3, 4, 5], Struwe [41], de Figueiredo-Miyagaki-Ruf [12], JM do Ó [13], de Figueiredo- do Ó-Ruf [11], YX Li [26, 27], Lu-Yang [33, 34], Lam-Lu [19, 21, 22] and the references therein Now, let us discuss the Moser-Trudinger type inequalities on the Heisenberg group For some notations and preliminaries about the Heisenberg group, see the next section In the setting of the Heisenberg group, although the rearrangement argument is not available, Cohn and the second author of this paper [9] can still set up a sharp Moser-Trudinger inequality for bounded domains on the Heisenberg group: Theorem A Let H = H n be a n dimensional Heisenberg group, = 2n + 2, = / 1), and α = 2π n Γ )Γ 2 )Γ 2 ) 1 Γn) 1) 1 Then there exists a constant C depending only on such that for all H n, <, 1 sup expα uξ) )dξ C < 11) u W 1, ), H u L 1

3 Subelliptic equations with critical growth in the Heisenberg group H n 661 If α is replaced by any larger number, the integral in 11) is still finite for any u W 1, H), but the supremum is infinite It is clear that when =, Theorem A is not meaningful In the case, we have the following version of the Moser-Trudinger type inequality see [1]): Theorem B Let α be such that α = α /c Then for any pair, α satisfying <, < α α, and α α + 1, there holds 1 { sup ) exp α u / 1) u W 1, H) 1 ρ ξ) S 2 α, u) } < 12) where H S 2 α, u) = 2 k= α k k! u k/ 1) When α α + > 1, the integral in 12) is still finite for any u W1, H), but the supremum is infinite if further α α + > 1 Here, c is defined as follows: Let u : H R be a nonnegative function in W 1, H), and u be the decreasing rearrangement of u, namely where u ξ) := sup {s :ξ {u > s} } {u > s} = B r = {ξ : ρ ξ) r} such that {u > s} = B r It is known from a result of Manfredi and V Vera De Serio [36] that there exists a constant c 1 depending only on such that H u dξ c H u dξ for all u W 1, H) Thus we can define { c = inf c 1/ 1) : H u dξ c H H H H } H u dξ, u W 1, H) 1 13) We notice that in Theorem B, we cannot exhibit the best constant α 1 ) due to the loss of the non-optimal rearrangement argument in the Heisenberg group Nevertheless, the first two authors recently used a completely different but much simpler approach, namely a rearrangementfree argument, to set up the sharp Moser-Trudinger type inequality on Heisenberg groups in [24] Moreover, we have developed in [25] a rearrangement-free method to establish the sharp Adams and singular Adams inequalities on high order Sobolev spaces W m, n m R n ) with arbitrary orders including fractional orders) This extends those results in [4] and [2, 23] in full generality The main result on sharp Moser-Trudinger inequality on the Heisenberg group proved in [24] is as follows Theorem C Let τ be any positive real number Then for any pair, α satisfying < and < α α 1 ), there holds 1 { sup ) exp α u / 1) u 1,τ 1 ρ ξ) S 2 α, u) } < 14) H When α > α 1 ), the integral in 14) is still finite for any u W1, H), but the supremum is infinite Here [ ] 1/ u 1,τ = H u + τ u H H

4 662 N Lam, G Lu, H Tang In this paper, we will prove the critical singular Moser-Trudinger inequality on bounded domains see Lemma 41) and study a class of partial differential equations of exponential growth by using the Moser-Trudinger type inequalities on the Heisenberg group More precisely, we consider the existence of nontrivial weak solutions for the nonuniformly subelliptic equations of sub-laplacian type of the form: div H a ξ, H u ξ))) = f ξ,uξ)) ρξ) u W 1, )\ {} u in where is a bounded domain with smooth boundary in H, + εhξ) in a ξ, τ) c h ξ) + h 1 ξ) τ 1) NU) for any τ in R 2 and ae ξ in, h L ) and h Lloc ), <,, f : R R behaves like exp α u ) when u, h W 1, ) ), h and ε is a positive parameter The main features of this class of problems are that it involves critical growth and the nonlinear operator sub-laplacian type In spite of a possible failure of the Palais-Smale PS) compactness condition, in this article we apply minimax method, in particular, the mountain-pass theorem to obtain the weak solution of NU) in a suitable subspace of W 1, ) Moreover, in the case of sub-laplacian, ie, a ξ, H u) = H u 2 H u, we will apply minimax methods, more precisely, the mountain-pass theorem combined with minimization and the Ekeland variational principle, to obtain the existence of at least two weak solutions to the nonhomogeneous problem div H H u 2 H u ) = f ξ,u) ρξ) u W 1, )\ {} u in + εhξ) in NH) Our paper is organized as follows: In Section 2, we give some notations and preliminaries about the Heisenberg group We also provide the assumptions on the nonlinearity f in this section We will discuss the variational framework and state our main results in Section 3 In Section 4, we proof critical singular Moser-Trudinger inequality on bounded domains and also, some basic lemmas that are useful in our paper We will investigate the existence of nontrivial solution to Eq NU) Theorem 31) in Section 5 The last section Section 6) is devoted to the study of multiplicity of solutions to equation NH) Theorems 32 and 33) 2 Preliminaries and assumptions First, we provide some notations and preliminary results Let H n = R 2n R be the n dimensional Heisenberg group Recall that the Heisenberg group H n is the space R 2n+1 with the noncommutative law of product x, y, t) x, y, t ) = x + x, y + y, t + t + 2 y, x x, y )), where x, y, x, y R n, t, t R, and, denotes the standart inner product in R n The Lie algebra of H n is generated by the left-invariant vector files T = t, X i = + 2y i x i t, Y i = 2x i, i = 1,, n y i t

5 Subelliptic equations with critical growth in the Heisenberg group H n 663 We will fix some notations: z = x, y) R 2n, ξ = z, t) H n, ρ ξ) = z 4 + t 2) 1/4, where ρ ξ) denotes the Heisenberg distance between ξ and the origin Denote H = H n, = 2n + 2, α = σ 1/ 1), σ = ρz,t)=1 z dµ We now use H u to express the norm of the subelliptic gradient of the function u : H R: n H u = Xi u) 2 + Y i u) 2) 1/2 i=1 Now, we will provide conditions on the nonlinearity of Eq NU) and NH) Motivated by the Moser-Trudinger inequalities Theorems A and B), we consider here the maximal growth on the nonlinear term f ξ, u) which allows us to treat EqNU) and NH) variationally in a suitable subspace of W 1, ) We assume that f : R R is continuous, f ξ, ) = and f behaves like exp α u / 1)) as u More precisely, we assume the following growth conditions on the nonlinearity f ξ, u): f ξ,s) f ξ,s) f 1)There exists α > such that lim s exp α s ) = uniformly on ξ, α > α and lim = uniformly on ξ, α < α f 2) There exist constant R, M > such that for all ξ and s R, Fξ, s) = s f ξ, τ)dτ M f ξ, s) Since we are interested in nonnegative weak solutions, it is convenient to define We note that conditions f 1), f 2) imply that: a) Fξ, s), ξ, s) R b) There is a positive constant C such that f ξ, u) = for all ξ, u), ] s R, ξ : Fξ, s) C exp ) 1 u M f 3) There exists p > and s 1 > such that for all ξ and s > s 1, < pfξ, s) sfξ, s) s exp α s ) Let A be a measurable function on R 2 such that Aξ, ) = and aξ, τ) = Aξ,τ) τ is a Caratheodory function on R 2 Assume that there are positive real numbers c, c 1, k, k 1 and two nonnegative measurable functions h, h 1 on such that h 1 Lloc ), h L / 1) ), h 1 ξ) 1 for ae ξ in and the following condition holds: A1) aξ, τ) c h ξ) + h 1 ξ) τ 1), τ R 2, aeξ A2) c 1 τ τ 1 aξ, τ) aξ, τ 1 ), τ τ 1, τ, τ 1 R 2, aeξ A3) aξ, τ) τ A ξ, τ), τ R 2, aeξ A4) A ξ, τ) k h 1 ξ) τ, τ R 2, aeξ Then A verifies the growth condition: A ξ, τ) c h ξ) τ + h 1 ξ) τ ), τ R 2, aeξ 25) For examples of A, we can consider A ξ, τ) = hξ) τ where h L loc )

6 664 N Lam, G Lu, H Tang 3 Variational framework and main results We introduce some notations: E = { u W 1, ) : h 1ξ) H u dξ < } 1 k M = lim exp k t t )) dt d is the radius of the largest open ball centered at contained in We notice that M is well-defined and is a real number greater than or equal to 2 see [13]) Under our conditions, we can see that E is a reflexive Banach space when endowed with the norm 1/ u = h 1 ξ) H u dξ) Furthermore, since h 1 ξ) 1 for ae ξ in : u λ 1 ) = inf : u E \ {} > for any < 36) u ρξ) dξ Now, from f 1), we obtain for all ξ, u) R, f ξ, u), F ξ, u) b 3 exp α 1 u / 1)) for some constants α 1, b 3 > Thus, by the Moser-Trudinger type inequalities, we have F ξ, u) L 1 ) for all u W 1, ) Define the functional E, T, J ε : E R by E u) = Aξ, Hu)dξ Tu) = Fξ,u) dξ ρξ) J ε u) = E u) Tu) ε hudξ then the functional J ε is well-defined Moreover, J ε is a C 1 functional on E with f ξ, u)v DJ ε u) v = a ξ, H u) H vdξ ρ ξ) dξ ε hvdξ, u, v E We next state our main results Theorem 31 Suppose that f1)-f3) are satisfied Furthermore, assume that f4) Fξ, s) lim sup s + k s < λ 1) uniformly in ξ Then there exists ε 1 > such that for each < ε < ε 1, NU) has a weak solution of mountain-pass type Theorem 32 In addition to the hypotheses in Theorem 31, assume that f5) lim sfξ, s) exp α s / 1)) > s ) 1 r 1 Mα 1 α 1 ω 1 uniformly on Then, there exists ε 2 >, such that for each < ε < ε 2, problem NH) has at least two weak solutions and one of them has a negative energy In the case where the function h does not change sign, we have Theorem 33 Under the assumptions in Theorems 31 and 32, if hξ) hξ) ) ae, then the solutions of problem NH) are nonnegative nonpositive)

7 Subelliptic equations with critical growth in the Heisenberg group H n Some lemmas First, we will prove the following critical singular Moser-Trudinger inequality: Lemma 41 The critical singular Moser-Trudinger inequality) Let H = H n be a n dimensional Heisenberg group, H n, <, = 2n + 2, = / 1), <, and α = σ 1/ 1), σ = ρz,t)=1 z dµ Then there exists a constant C depending only on and such that 1 expα 1 ) uξ) )dξ sup u W 1, 1 ρ ξ) C < ), H u L 1 If α 1 ) is replaced by any larger number, then the supremum is infinite Recall in [9] that for < α <, we will say that a non-negative function g on H is a kernel of order α if g has the form gξ) = ρ ξ) α g ξ ) where ξ = ξ ρξ) is a point on the unit sphere We are also assuming that for every δ > and < M < there are constants Cδ, M) such that Σ δ M g ξ sζ ) 1) g ξ ) ds s dµ ξ ) Cδ, M) for all ζ Σ = {ξ H : ρ ξ) = 1}, where Σ δ is the subset of the unit sphere given by Σ δ = { ξ Σ : δ g ξ ) δ 1} Note that we will choose g ξ) = z 1 in the proof of Lemma 41 First, we will prove the following ξ 2 2 result: Lemma 42 Suppose g is an allowed kernel of order α, pα =, H, <, f L p and < Let Ag, p) = A α g) = g ξ ) p dµ Here p = p/p 1) Then there exists a constant C depending only on and such that [ exp Ag, p) ) ) p ] 1 f gξ) f 1 p 1 ρ ξ) dµ C Proof Set Then u ξ) = f gξ) φs) = 1/p f e s )e s/p f ξ) p dξ = = Σ f t) p dt φs) p ds By the Hardy-Littlewood inequality, the O Neil s lemma, noting that with hξ) = 1, then h t) = ρξ) ), where C is the volume of the unit ball, we have C t

8 666 N Lam, G Lu, H Tang C exp A g, p) 1 ρ ξ) e Ag,p) ) 1 u t) p t = C 1 ) uξ) p ) dt dξ [ exp 1 ) A g, p) u e s) p 1 ) ] s ds C 1 e ) s p exp 1 p e s ) 1 p f z)dz + f z)z 1 p dz e s exp [ ) ] ds 1 s ) s p exp 1 φw)e w pes/p p dw + φw) = C 1 s exp [ ) ] ds 1 s = C 1 [ ] exp F 1 ) s) ds where F 1 ) s) = as, t) = 1 ) t 1 ) 1 for < s < t for t < s < for < s pe t s)/p p as, t)φs)ds, Using Lemma 31 in [23], we get our desired result Proof of Lemma 41) Now, using Lemma 42, noting that by Theorem 12 in [9], we have that u ξ) σ ) 1 H u g ξ), where g ξ) = z 1 ξ 2 2, we can derive the result of Lemma 41 Note that the sharpness of the constant α 1 ) comes from the test functions in [9]

9 Subelliptic equations with critical growth in the Heisenberg group H n 667 Using the critical singular Moser-Trudinger inequality, we can prove the following two lemmas see [14] and [1]): Lemma 43 For κ >,q > and u M = M, κ) with M sufficiently small, we have exp κ u / 1)) Proof By Holder inequality and Lemma 41, we have ρ ξ) u q dξ C, κ) u q exp κ u / 1)) ρ ξ) u q dξ exp κr u ) u ) u ρ ξ) r dξ ) 1/r C) u r q dξ 1/r u r q dξ if r > 1 sufficiently close to 1 Here r = r/r 1) By the Sobolev embedding, we get the result ) 1/r Lemma 44 If u E and u N with N sufficiently small κn for some s > 1 Proof The proof is similar to Lemma 43 exp κ u / 1)) ρ ξ) v dξ C, M, κ) v s We also have the following lemma for Euclidean case, see [31]): < α 1 ) ), then Lemma 45 Let {w k } W 1, ), H w k L ) 1 If w k w weakly and almost everywhere, H w k H w almost everywhere, then exp{α w k / 1) } is bounded in L 1 ) for ρξ) < α < 1 ) α 1 H w 1/ 1) L )) Proof Using Brezis-Lieb Lemma in [7], we deduce that H w k L ) Hw k H w L ) Hw L ) Thus for k large enough and δ > small enough: < α 1 + δ) H w k H w / 1) < 1 ) α L ) 47) Now, noting that for some Cδ) > : w k / 1) 1 + δ 2 ) w k w / 1) + Cδ) w / 1)

10 668 N Lam, G Lu, H Tang by Holder inequality, we have exp { α w k / 1)} ρ ξ) dξ exp { α1 + δ 2 ) w k w / 1) + αcδ) w / 1)} ρ ξ) dξ exp { αq1 + δ 2 ) w k w / 1)} exp ρ ξ) dξ 1/q { αq Cδ) w / 1)} ρ ξ) dξ exp { α1 + δ) w k w / 1)} 1/q C ρ ξ) dξ where we choose q = 1+δ 1+ δ 2 is bounded in L 1 ) and q = q q 1 Now, by 47) and Lemma 41, we have that exp{α w k / 1) } ρξ) 5 The existence of solution for the problem NU) The existence of nontrivial solution to Eq NU) will be proved by a mountain-pass theorem without a compactness condition such like the one of the PS) type This version of the mountain-pass theorem is a consequence of the Ekeland s variational principle First, we will check that the functional J ε satisfies the geometric conditions of the mountain-pass theorem Lemma 51 Suppose that f 1) and f 4) hold Then there exists ε 1 > such that for < ε < ε 1, there exists ρ ε > such that J ε u) > if u = ρ ε Furthermore, ρ ε can be chosen such that ρ ε as ε Proof From f 4), there exist τ, δ > such that u δ implies Fξ, u) k λ 1 ) τ) u 58) for all ξ Moreover, using f 1) for each q >, we can find a constant C = Cq, δ) such that for u δ and ξ From 58) and 59) we have Fξ, u) C u q exp κ u / 1)) 59) Fξ, u) k λ 1 ) τ) u + C u q exp κ u / 1)) for all ξ, u) R Now, by A4), Lemma 43, 36) and the continuous embedding E L ), we obtain J ε u) k u k λ 1 ) τ) u ρ ξ) dξ C u q ε h u k 1 λ ) 1 ) τ) u C u q ε h λ 1 ) u Thus J ε u) u [k 1 λ ) ] 1 ) τ) u 1 C u q 1 ε h λ 1 ) 51) ) Since τ > and q >, we may choose ρ > such that k 1 λ 1 ) τ) λ 1 ) ρ 1 Cρ q 1 > Thus, if ε is sufficiently small then we can find some ρ ε > such that J ε u) > if u = ρ ε and even ρ ε as ε 1/q

11 Subelliptic equations with critical growth in the Heisenberg group H n 669 Lemma 52 There exists e E with e > ρ ε such that J ε e) < inf u =ρ ε J ε u) Proof Let u E \ {}, u with compact support = suppu) By f 2) and f 3), we have that for p >, there exists a positive constant C > such that Then by 25), we get J ε γu) Cγ s, ξ : F ξ, s) cs p d 511) h ξ) H u dξ + Cγ u Cγ p u p dξ + C + εγ ρ ξ) hudξ Since p >, we have J ε γu) as γ Setting e = γu with γ sufficiently large, we get the conclusion In studying this class of sub-elliptic problems involving critical growth, the loss of the PS) compactness condition raises many difficulties In the following lemmas, we will analyze the compactness of PS) sequences of J ε Lemma 53 Let u k ) E be an arbitrary PS) sequence of J ε, ie, J ε u k ) c, DJ ε u k ) in E as k Then there exists a subsequence of u k ) still denoted by u k )) and u E such that f ξ,u k ) ρξ) strongly in Lloc 1 ) H u k ξ) H uξ) almost everywhere in a ξ, H u k ) a ξ, H u) weakly in L / 1) ) ) 2 loc u k u weakly in E f ξ,u) ρξ) Furthermore u is a weak solution of NU) In order to prove this lemma, we need the following two lemmas that can be found in [1], [13], [14] and [32]: Lemma 54 Let B r ξ ) be a Heisenberg ball centered at ξ ) with radius r Then there exists a positive ε depending only on such that sup B r ξ exp ε u / 1)) dξ C ) Brξ ) Hu dξ 1, Brξ ) udξ= 1 for some constant C depending only on Lemma 55 Let u k ) in L 1 ) such that u k u in L 1 ) and let f be a continuous function Then f ξ,u k ) f ξ,u) in L 1 ), provided that f ξ,u kξ)) L 1 ) k and f ξ,u k ξ))u k ξ) dξ C ρξ) ρξ) ρξ) ρξ) 1 Now we are ready to prove Lemma 53 Proof The proof is similar to Lemma 34 in [1] For the completeness, we sketch the proof here By the assumption, we have Fξ, u k ) Aξ, H u k )dξ ρ ξ) dξ ε hu k dξ c 512) B r ξ )

12 67 N Lam, G Lu, H Tang and a ξ, H u k ) H vdξ f ξ, u k )v ρ ξ) dξ ε hvdξ τ k v 513) for all v E, where τ k as k Choosing v = u k in 513) and by A3), we get f ξ, u k )u k ρ ξ) dξ + ε hu k dξ A ξ, H u k ) τ k u k This together with 512), f 3) and A4) leads to ) p 1 u k C 1 + u k ) and hence u k is bounded and thus f ξ, u k )u k ρ ξ) dξ C, Fξ, u k ) dξ C 514) ρ ξ) Note that the embedding E L q ) is compact for all q 1, by extracting a subsequence, we can assume that u k u weakly in E and for almost all ξ Thanks to Lemma 55, we have f ξ, u k ) ρ ξ) f ξ, u) ρ ξ) in L1 ) 515) Now, similar to that in [1], up to a subsequence, we define an energy concentration set for any fixed δ >, { Σ δ = ξ : lim lim uk r B + H u k ) } dξ δ r ξ) Since u k ) is bounded in E, Σ δ must be a finite set For any ξ Σ δ, there exist r :< r < dist ξ, Σ δ ) such that lim uk + H u k ) dξ < δ B r ξ ) so for large k : B r ξ ) uk + H u k ) dξ < δ 516) By results in [1], we can prove that f ξ, u k ) u k u B r ξ ) ρ ξ) dξ 517) f ξ, u k ) ρ ξ) /q 1 1/q L q ρ ξ) u k u L C u q s k u L q s L s and for any compact set K \ Σ δ, lim K f ξ, u k ) u k f ξ, u) u ρ ξ) dξ = 518) So now, we will prove that for any compact set K \ Σ δ, H u k H u dξ = 519) lim K

13 Subelliptic equations with critical growth in the Heisenberg group H n 671 It is enough to prove for any ξ \ Σ δ, and r given by 516), there holds H u k H u dξ = 52) lim B r/2 ξ ) For this purpose, we take φ C B r ξ )) with φ 1 and φ = 1 on B r/2 ξ ) Obviously φu k is a bounded sequence Choose v = φu k and v = φu in 513), we have: φ a ξ, H u k ) a ξ, H u)) H u k H u) dξ B r ξ ) a ξ, H u k ) H φ u u k ) dξ + B r ξ ) B r ξ ) φa ξ, H u) H u H u k ) dξ + φh u k u) dξ + τ k φu k + τ k φu + ε B r ξ ) B r ξ ) φ u k u) f ξ, u k) ρ ξ) dξ Note that by Holder inequality and the compact embedding of E L ), we get a ξ, H u k ) H φ u u k ) dξ = 521) lim B r ξ ) Since H u k H u and u k u, there holds lim B r ξ ) φa ξ, H u) H u H u k ) dξ = and lim φh u k u) dξ = 522) B r ξ ) The Holder inequality and 517) implies that φ u k u) f ξ, u k ) dξ = So we can conclude that lim B r ξ ) lim B r ξ ) φ a ξ, H u k ) a ξ, H u)) H u k H u) dξ = and hence we get 52) by A2) So we have 519) by a covering argument Since Σ δ is finite, it follows that H u k converges to H u almost everywhere This immediately implies, up to a subsequence, a ξ, H u k ) a ξ, H u) weakly in L / 1) ) ) 2 loc Let k tend to infinity in 513) and combine with 515), we obtain This completes the proof of the Lemma 51 The proof of Theorem 31 DJ ε u), h = h C ) Proposition 51 Under the assumptions f1)-f4), there exists ε 1 > such that for each < ε < ε 1, the problem NU) has a solution u M via mountain-pass theorem Proof For ε sufficiently small, by Lemma 51 and 52, J ε satisfies the hypotheses of the mountainpass theorem except possibly for the PS) condition Thus, using the mountain-pass theorem without the PS) condition, we can find a sequence u k ) in E such that J ε u k ) c M > and DJ ε u k ) where c M is the mountain-pass level of J ε Now, by Lemma 53, the sequence u k ) converges weakly to a weak solution u M of NU) in E Moreover, u M since h

14 672 N Lam, G Lu, H Tang 6 The multiplicity results to the problem NH) In this section, we study the problem NH) Note that Eq NH) is a special case of the problem NU) where A ξ, τ) = τ, 1/ u = H u dξ) As a consequence, there exists a nontrivial solution of standard mountain-pass type as in Theorem 31 Now, we will prove the existence of the second solution Lemma 61 There exists η > and v E with v = 1 such that J ε tv) < for all < t < η In particular, inf u η J εu) < Proof Let v E be a solution of the problem { divh H v 2 H v ) = h in v = on Then, for h, we have hv = v > Moreover, d dt J εtv) = t 1 v f ξ, tv) v ρ ξ) dξ ε hvdξ for t > Since f ξ, ) =, by continuity, it follows that there exists η > such that d dt J εtv) < for all < t < η and thus J ε tv) < for all < t < η since J ε ) = Next, we define the Moser Functions see [1, 19]): m k ξ, r) = 1 σ 1/ ) log k 1)/ if ρξ) r k log r if r k ρξ) r ρξ) log k) 1/ if ρξ) r Using the fact that H ρξ) = z ρξ) where ξ = z, t), we can conclude that m k, r) W 1, ), the support of m k ξ, r) is the ball B r, and H m k ξ, r) dξ = 1 623) Lemma 62 Suppose that f1)-f5) hold Then there exists k N such that { t max t Proof Let r > and > be such that } F ξ, tm k ) ρ ξ) dξ < 1 lim sf ξ, s) exp α s ) > s α α ) 1 ) 1 r 1 Mα 1 α 1 ω 1

15 Subelliptic equations with critical growth in the Heisenberg group H n 673 uniformly for almost every ξ and B, r) Let m k ξ) = m k ξ, r) Then we have Suppose, by contradiction, that for all k we get { t max t For each k, we can find t k > such that Thus t k From Fξ, u), we obtain Since at t = t k we have it follows that t k m k W 1, B r ) and m k = 1 } F ξ, tm k ) ρ ξ) dξ 1 { F ξ, t k m k ) t ρ ξ) dξ = max t t k = F ξ, t k m k ) ρ ξ) dξ 1 t k d t dt α α α α ) 1 } F ξ, tm k ) ρ ξ) dξ α α ) 1 ) 1 624) ) F ξ, tm k ) ρ ξ) dξ = t k m k f ξ, t k m k ) t k m k f ξ, t k m k ) ρ ξ) dξ = ρξ) r ρ ξ) dξ 625) Using hypothesis f 5), given τ > there exists R τ > such that for all u R τ and ρ ξ) r, we have ufξ, u) τ) exp α u / 1)) 626) From 625) and 626), for large k, we obtain Thus, setting we have t k τ) ρξ) r k = τ) ω 1 r k = τ) ω 1 r exp exp α t k m k / 1)) ρ ξ) dξ ) exp α t / 1) k α σ 1/ 1) log k ) t / 1) α k log k log k + log k L k = α log k t / 1) α k log t k ) log k 1 τ) ω 1 r exp L k Consequently, the sequence t k ) is bounded Moreover, by 624) and t k τ) ω 1 r exp α t / 1) k ) α log k )

16 674 N Lam, G Lu, H Tang it follows that Set t k α α A k = {ξ B r : t k m k R τ } and B k = B r \ A k ) 1 627) From 625) and 626) we have t k exp α t k m k / 1)) τ) ρξ) r ρ ξ) dξ + exp α t k m k / 1)) τ) B k t k m k f ξ, t k m k ) B k ρ ξ) ρ ξ) dξ 628) Notice that m k ξ) and the characteristic functions χ Bk 1 for almost everywhere ξ in B r Therefore the Lebesgue dominated convergence theorem implies t k m k f ξ, t k m k ) B k ρ ξ) dξ and Moreover, using we have ρξ) r = ρξ) r B k exp α t k m k / 1)) ρ ξ) dξ ω 1 r t k exp α t k m k / 1)) ρ ξ) dξ ρξ) r/k exp α m k / 1)) ρ ξ) dξ exp α m k / 1)) ρ ξ) dξ + α α ) 1, r/k ρξ) r dξ exp α m k / 1)) ρ ξ) dξ and ρξ) r/k exp α m k / 1)) ρ ξ) dξ = ρξ) r/k = ω 1 [ ] exp α σ 1/ 1) log k dξ ) k ) r k = ω 1 r Now, using the change of variable ζ = log r s ) log k

17 Subelliptic equations with critical growth in the Heisenberg group H n 675 by straightforward computation, we have r/k ρξ) r = ω 1 r log k exp α m k / 1)) ρ ξ) dξ 1 exp [ ) log k ζ / 1) ζ )] dζ which converges to ω 1 r M as k Finally, taking k in 628) and using 627), we obtain ) α 1 τ) ω 1r M which implies that ) 1 α 1, 1 r Mα 1 ω 1 but it is a contradiction and the proof is complete α Corollary 61 Under the hypotheses f1)-f5), if ε is sufficiently small then { t max J ε tm k ) = max t t } F ξ, tm k ) ρ ξ) dξ t εhm k dξ < 1 Proof Since εhm kdξ ε h, taking ε sufficiently small, the result follows Note that we can conclude by inequality 51) and Lemma 61 that α α ) 1 < c ε = inf u ρ ε J ε u) < 629) Next, we will prove that this infimum is achieved and generate a solution In order to obtain convergence results, we need to improve the estimate of Lemma 62 Corollary 62 Under the hypotheses f1)-f5), there exist ε 2, ε 1 ] and u W 1, ) with compact support such that for all < ε < ε 2, J ε tu) < c ε + 1 α α ) 1 for all t ε Proof It is possible to raise the infimum c ε by reducing ε By Lemma 51, ρ ε Consequently, ε c ε Thus there exists ε 2 > such that if < ε < ε 2 then, by Corollary 61, we have max t J ε tm k ) < c ε + 1 Taking u = m k W 1, ), the result follows Lemma 63 If u k ) is a PS) sequence for J ε at any level with α α ) 1 lim inf u k < N 63) with N sufficiently small, then u k ) possesses a subsequence which converges strongly to a solution u of NH)

18 676 N Lam, G Lu, H Tang Proof Extracting a subsequence of u k ) if necessary, we can suppose that lim inf u k = lim u k < N By Lemma 53, we have that u k u where u is a weak solution of NH) Writing u k = u + w k, it follows that w k in E Thus w k in L q ) for all 1 q < Using the Brezis-Lieb Lemma in [7], we have u k = u + w k + o k 1) 631) We claim that f ξ, u k )u ρ ξ) dξ f ξ, u )u ρ ξ) dξ 632) In fact, using u E, given τ >, there exists ϕ C ) such that ϕ u < τ We have that f ξ, u k )u f ξ, u )u ρ ξ) dξ ρ ξ) dξ f ξ, u k ) f ξ, u ) + ϕ ρ ξ) dξ + suppϕ f ξ, u k ) u ϕ) ρ ξ) dξ f ξ, u ) u ϕ) ρ ξ) dξ Since DJ ε u k )u ϕ) τ k u ϕ with τ k, we have f ξ, u k ) u ϕ) ρ ξ) dξ τ k u ϕ + H u k 1 u ϕ + εh u ϕ C u ϕ < Cτ, where C is independent of k and τ Similarly, using that DJ ε u )u ϕ) =, we have f ξ, u ) u ϕ) ρ ξ) dξ < Cτ Since f ξ,u k) f ξ,u ) strongly in L 1 ) and by the previous inequalities, we conclude that ρξ) ρξ) lim f ξ, u k )u ρ ξ) dξ f ξ, u )u ρ ξ) dξ < 2Cτ and this shows the convergence 632) because τ is arbitrary From 631) and 632), we can write that is DJ ε u k ) u k = DJ ε u ) u + w k w k = f ξ, u k ) w k ρ ξ) dξ + o k 1) f ξ, u k ) w k ρ ξ) dξ + o k 1) From f 1), Lemma 41, Lemma 44 and the compact embedding E L t ) for t 1, we have f ξ, u k ) w k ρ ξ) dξ from which w k and the result follows

19 Subelliptic equations with critical growth in the Heisenberg group H n Proof of Theorem 32 The proof of the existence of the second solution of NH) follows by a minimization argument and Ekeland s variational principle Proposition 61 There exists ε 2 > such that for each ε with < ε < ε 2, Eq NH) has a minimum type solution u with J ε u ) = c ε <, where c ε is defined in 629) Proof Let ρ ε, N be as in Lemma 51 and Lemma 63 Note that we can choose ε 2 > sufficiently small such that ρ ε < N Since B ρε is a complete metric space with the metric given by the norm of E, convex and the functional J ε is of class C 1 and bounded below on B ρε, by the Ekeland s variational principle there exists a sequence u k ) in B ρε such that Observing that J ε u k ) c ε = inf u ρ ε J ε u) and DJ ε u k ) u k ρ ε < N by Lemma 63, it follows that there exists a subsequence of u k ) which converges to a solution u of NH) Therefore, J ε u ) = c ε < Remark 61 By Corollary 62, we can conclude that < c M < c ε + 1 α α ) 1 Proposition 62 If ε 2 > is enough small, then the solutions of NH) obtained in Propositions 51 and 61 are distinct Proof By Proposition 51 and 61, there exist sequences u k ), v k ) in E such that and u k u, J ε u k ) c ε <, DJ ε u k ) u k v k u M, J ε v k ) c M >, DJ ε v k ) v k, H v k ξ) H u M ξ) ae Now, suppose by contradiction that u = u M Then we have that v k u As in the proof of Lemma 53 we obtain f ξ, v k ) ρ ξ) f ξ, u ) ρ ξ) in L 1 ) 633) From this, we have by f 2), f 3) and the Generalized Lebesgue s Dominated Convergence Theorem: Fξ, v k ) ρ ξ) Fξ, u ) ρ ξ) in L 1 ) Now, we have that J ε u k ) = 1 u k Fξ, u ) ρ ξ) ε hu + o1) = c ε + o1) and J ε v k ) = 1 v k Fξ, u ) ρ ξ) ε hu + o1) = c M + o1)

20 678 N Lam, G Lu, H Tang which implies u k v k c ε c M ) < 634) Also, since u k ), v k ) are both bounded Palais-Smale sequences, we have DJ ε u k ) u k = u k f ξ, u k ) ρ ξ) u k ε hu k DJ ε v k ) v k = v k f ξ, v k ) ρ ξ) v k ε hv k and then uk v k ) [ f ξ, uk ) ρ ξ) u k v k ) + ε [h u k u ) h v k u )] ) ] f ξ, uk ) f ξ, v k ) v k ρ ξ) tends to as k But it is clear that [h u k u ) h v k u )] 635) since h W 1, ) ) and uk u, v k u in W 1, ) Also, notice that f ξ, u k ) u ρ ξ) k v k ) 636) Indeed, let w k = v k u Thus w k and lim w k > since w k Again, using Holder inequality, Lemma 41 and Theorem A, note that u k is small, we have f ξ, u k ) ) q u ρ ξ) k v k ) f ξ, uk ) 1/q 1/q dξ ρ ξ) u k v k ) q So, it remains to show that C u k v k q f ξ, uk ) f ξ, v k ) ρ ξ) The left hand side of 637) can be written as ) f ξ, uk ) f ξ, v k ) ρ ξ) u dξ + ) v k dξ 637) f ξ, uk ) f ξ, v k ) ρ ξ) ) w k dξ Arguing as in Lemma 63, we can conclude that ) f ξ, uk ) f ξ, v k ) ρ ξ) u dξ 638) Now, again by Holder inequality, Theorem A and the Sobolev embedding, we get f ξ, u k ) ρ ξ) w kdξ C w k q 639) So now, we just need to prove f ξ, u k ) ρ ξ) w kdξ 64)

21 Subelliptic equations with critical growth in the Heisenberg group H n 679 Indeed, for large k, we have c M c ε = J ε v k ) J ε u k ) + o1) = 1 v k 1 u + o1) < 1 α α ) 1 Therefore, we can find s > 1 sufficiently close to 1 such that for large k, Thus, ) s v k u α 1 < sα sα v k < α s Define V k = v k v k Thus V k = 1, V k V = information that w k s, we have f ξ, v k ) ρ ξ) w kdξ C 1 u ) 1/ 1) v k u lim v k and V < 1 By Lemma 45 and the [ exp sα v k v k ρ ξ) s v k ] 1/s w k s and then we have 64) From 635), 636), 637), 638), 639) and 64) we have u k v k which is a contradiction to 634) The proof is completed now 62 Proof of Theorem 33 Corollary 63 There exists ε 3 > such that if < ε < ε 3 and hξ) for all ξ, then the weak solutions of NH) are nonnegative Proof Let u be a weak solution of NH), that is, H u 2 H u H vdξ f ξ, u) v ρ ξ) dξ εhvdξ = for all v E Taking v = u E and observing that f ξ, u ξ)) u ξ) = ae, we have u = εhu dξ E Consequently, u = u + H Acknowledgements: The authors wish to thank the referee for his/her many helpful comments which have improved the exposition of the paper

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